Question: Which integral gives the length of the graph of $y=e^{2x}$ between $x=-2$ and $x=0$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_{-2}^0\sqrt{1+4e^{4x}}~dx$ (Choice B) B $ \int_{-2}^0\sqrt{1+2e^{2x}}~dx$ (Choice C) C $ \int_{-2}^0\sqrt{1+e^{4x}}~dx$ (Choice D) D $ \int_{-2}^0\sqrt{1+e^{2x}}~dx$
Answer: Recall that the formula for arc length of $~f(x)~$ over $~[a, b]~$ is $ L=\int_a^b\sqrt{1+\big[f\,^\prime(x)\big]^2}~dx\,$. First calculate $f\,^\prime(x)$. $ f\,^\prime(x)=e^{2x}\cdot2=2e^{2x}$ Next, use the formula above to write the integral expression that gives the arc length in question and simplify. $ L= \int_{-2}^0\sqrt{1+\big(2e^{2x}\big)^2}~dx=\int_{-2}^0\sqrt{1+4e^{4x}}~dx$